Abstract

It was proved (Bessy et al., 2010) that for , a tournament with minimum semidegree at least contains at least r vertex-disjoint directed triangles. It was also proved (Lichiardopol, 2010) that for integers and , every tournament with minimum semidegree at least () contains at least r vertex-disjoint directed cycles of length . None information was given on these directed cycles. In this paper, we fill a little this gap. Namely, we prove that for and , every tournament of minimum outdegree at least contains at least vertex-disjoint strongly connected subtournaments of minimum outdegree . Next, we prove for tournaments a conjecture of Stiebitz stating that for integers and , there exists a least number such that every digraph with minimum outdegree at least can be vertex-partitioned into two sets inducing subdigraphs with minimum outdegree at least and at least , respectively. Similar results related to the semidegree will be given. All these results are consequences of two results concerning the maximum order of a tournament of minimum outdegree (of minimum semidegree ) not containing proper subtournaments of minimum outdegree (of minimum semidegree ).

1. Introduction and Definitions

Let be a digraph. is the vertex set of and the order of is the cardinality of . is the set of the arcs of . Two vertices and of are adjacent, if at least one of the ordered pairs and is an arc of . We say that a vertex is an outneighbor of a vertex (inneighbour of ) if (resp. ) is an arc of . is the set of the outneighbors of and is the set of the in-neighbors of . The cardinality of is the outdegree of and the cardinality of is the indegree of . When no confusion is possible, we omit the subscript . We denote by the minimum outdegree of and by the minimum indegree of . The minimum semidegree of is .

An oriented graph, is a digraph such that for any two distinct vertices and of , at most one of the couples and is an arc of .

A tournament is an oriented graph such that any two distinct vertices and of are adjacent. If and are subsets of , an arc from to is an arc with and . We denote by the number of the arcs from to .

It is known and easy to prove that if is the order of , then and for every vertex of .

For a subset of , is the subtournament induced by the vertices of . For a vertex of , is the subtournament induced by the vertices of distinct from .

For , a regular tournament of degree is a tournament with for every vertex of . It is known and easy to prove that the order of is .

A path or a cycle of a tournament always means a directed path or a directed cycle of and disjoint cycles means vertex-disjoint cycles. A triangle is a directed cycle consisting of three vertices

For distinct vertices and of , an -path is a directed path starting from and ending at . The tournament is said to be strongly connected, or briefly strong, if for any distinct vertices and , there exists an -path. It is well known (Camion Theorem) that a tournament is strong if and only if it contains a Hamiltonian cycle. The strong connectivity of is the smallest nonnegative integer such that there exists a subset of vertices of disconnecting . For , a -strong tournament is a tournament of strong connectivity at least .

If and are two vertex-disjoint subsets of , we say that dominates , if for every pair with and , is an arc of (which means that there is no arc from to ). If and are two tournaments with disjoint vertex sets, is the tournament whose vertex set is and whose arcs are those of and and the ordered pairs with and . It is known and easy to prove that a tournament is nonstrong if and only if there exists a partition of such that .

A minimum outdegree minimal tournament is a tournament such that every proper subtournament of has minimum outdegree at most . If , we say that is a minimum outdegree minimal tournament. Similarly, one can define the notion of minimum indegree minimal tournament. A minimum semidegree minimal tournament is a tournament such that every proper subtournament of has minimum semidegree at most . If , we say that is a minimum semidegree minimal tournament.

In a recent paper, Bessy et al. (see [2]) proved that for , a tournament with minimum outdegree and minimum indegree both greater or equal to contains at least vertex-disjoint directed triangles. In a more recent paper (see [3]), the author generalized this result, by proving that for given integers and , every tournament with minimum outdegree and minimum indegree both greater or equal to contains at least vertex-disjoint directed cycles of length . None information was given on these directed cycles. In this paper, we fill a little this gap. More precisely, we prove:

Theorem 1.1. For and for , every tournament of minimum outdegree at least contains at least vertex-disjoint strong subtournaments of minimum outdegree .

Concerning the minimum semidegree, we prove the following.

Theorem 1.2. For and for , every tournament of minimum semidegree at least contains at least vertex-disjoint subtournaments of minimum semidegree .

Conjecture 1. For given integers and , there exists a least number such that the vertices of any simple digraph with minimum outdegree at least can be partitioned into two sets inducing subdigraphs with minimum outdegree at least and at least , respectively.

A natural generalization given by Alon (Problem 1 in [5]) is as follows.

Conjecture 2. For an integer and for given positive integers , there exists a least number such that the vertices of any simple digraph with minimum outdegree at least can be partitioned into sets such that for .

It is easy to prove by induction that the existence of implies the existence of . In this paper, we prove that Stiebitz's conjecture is true for tournaments (with a supplementary constraint), namely, we prove the following.

Theorem 1.3. For given integers and , the vertices of any tournament with minimum outdegree at least can be partitioned into two sets inducing subtournaments and with strongly connected and of minimum outdegree at least and with of minimum outdegree at least .

This result will allow us to prove, for tournaments, the generalized conjecture of Stiebitz. Relatively to the minimum semidegree, we state the following.

Theorem 1.4. For given integers and , the vertices of any tournament with minimum semidegree at least can be partitioned into two sets inducing subtournaments and with of minimum semidegree at least and with of minimum semidegree at least .

Proof. Let be the smallest order of the subtournaments of having minimum outdegree at least . There exists a subtournament of of order and of minimum outdegree at least . Clearly, is a minimum outdegree minimal subtournament.

Concerning the order of a minimum outdegree minimal tournament, we state the following,

Theorem 2.3. For , if is a minimum outdegree minimal tournament of order , one has .

Proof. Let be the set of the vertices of of outdegree and let be its cardinality. For every vertex of , the tournament has minimum outdegree , and this means that has at least one in-neighbor in . Then, the number of the arcs from to is at least . On the other hand, the number of the arcs from to is exactly . It follows , hence . This implies , hence and since is an integer, we get , that is .

We note that . When , we get . It is easy to prove that there are four minimum outdegree 2 minimal tournaments: the regular tournament of order 5 (Figure 1), and three nonisomorphic tournaments , , and of order 6 (Figure 2). The outdegree sequence of is , and the outdegree sequence of the tournaments and is . We observe that and are not 2-connected (the two others yes).

Figure 1: Minimum outdegree 2 minimal tournament .

Figure 2: Minimum outdegree 2 minimal tournaments , , and .

We claim that for every integer the bound of Theorem 2.3 is reached. To be more precise, we claim that for every integer , there exists a minimum outdegree minimal tournament of order and of strong connectivity 1. We are going to prove this, by induction on . Clearly, the assertion is true for . Suppose that the assertion is true up to the row , . Let then be a minimum outdegree minimal tournament of order . Let be the tournament defined as follows.(i)The vertices of are the vertices of and additional vertices and .(ii)The arcs of are the arcs of , the couples , , the couples with , the couples with and and the arcs of an arbitrary tournament with as vertex set. It is easy to see that is a minimum outdegree minimal tournament of order and of strong connectivity 1 (because disconnects ). So the assertion is true for and consequently the result is proved.

In fact, there are minimum outdegree minimal tournaments of strong connectivity 1 of order smaller than . Indeed, if we take a regular tournament of degree , with the above construction, we get a minimum outdegree minimal tournament of connectivity 1 and of order . We think that is the minimum order of a minimum outdegree minimal tournament of connectivity 1.

By minimum outdegree critical tournament, we mean a tournament of minimum outdegree such that for every vertex of , the tournament has minimum out degree . It is clear that a minimum outdegree minimal tournament is minimum outdegree critical. By the way, we observe that in the proof of Theorem 2.3, we use only the fact that is minimum outdegree critical and as the obtained upper bound is reached, another proof using the fact that is minimum outdegree minimal cannot improve this upper bound. However, we claim that for , the notion of minimum outdegree minimal tournament does not coincide with the notion of minimum outdegree critical tournament. Indeed, let be a regular tournament of degree and consider a vertex of . We define the tournament in the following way.(i)The vertices of are the vertices of and the additional vertices , and of a directed triangle . (ii) is dominated by , is dominated by . To each vertex of we join exactly outneighbors in so that every vertex of has at least one in-neighbor in (this is possible when ).

has minimum outdegree and the vertices , and are of outdegree . It is easy to see that every vertex of is dominated by a vertex of . This means that is minimum outdegree critical. Manifestly, since is a proper subtournament of of minimum outdegree , is not minimum outdegree minimal. This corroborates our statement. However, as for minimum outdegree minimal tournaments, we have the following.

Theorem 2.4. For , any critically outdegree tournament is strong.

Proof. Suppose the opposite. Then there exists a partition of such that dominates . Then is a tournament of minimum outdegree at least , which implies . Consider a vertex of . An in-neighbor of is in and then its outdegree is greater than . This means that is a tournament of minimum outdegree , which is not possible. Consequently is strong.

For minimum semidegree minimal tournaments, the situation is a little different. Already, we observe that a minimum semidegree minimal tournament is not necessarily strong. Indeed, if is a minimum indegree minimal tournament and if is a a minimum outdegree minimal tournament, vertex disjoint with , it is easy to prove that is a nonstrong minimum semidegree minimal tournament. As for Theorem 2.2, we have the following.

Proof. Let be the smallest order of the subtournaments of having minimum semidegree at least . There exists a subtournament of of order and of minimum semidegree at least . Clearly, is a minimum semidegree minimal subtournament.

As regards the maximum order, we state the following.

Theorem 2.6. For , if is a minimum semidegree d minimal tournament of order and of minimum semidegree , we have .

Proof. We have and if , the theorem is proved. So, we may suppose . For every vertex of , the tournament has minimum semidegree . This means that has an in-neighbor of outdegree or an outneighbor of indegree (here “or” is not exclusive). Denote by the set of the vertices of of outdegree and by the set of the vertices of of indegree . Since , and are disjoint. Let and be the cardinalities of the sets and . If or if , as in the proof of Theorem 2.3, we get and then the theorem is proved. So, we may suppose and . Denote by the set of the vertices of not in and having at least one in-neighbor in , and by the cardinality of . Let and . Every possible vertex of has a least one outneighbor in .We have , and since , it follows that
Similarly, we get
From (2.1) and (2.2), we get by addition . It is easy to prove that . It follows , hence . This yields . This implies , hence and since is an integer, we get , that is .

Concerning the number of the vertices of with either outdegree or indegree , we have . When , we get , and this bound is reached. Indeed if is a critically indegree tournament of order 6 and if is a a critically outdegree tournament of order 6, then is a critically semidegree tournament of order 6.

Since , by Theorem 2.2, contains minimum outdegree minimal subtournaments. Let be the maximum number of vertex-disjoint minimum outdegree minimal subtournaments. Then there exist vertex-disjoint minimum outdegree minimal subtournaments , and by Theorem 2.3, these tournaments cover a set with at most vertices of . Suppose that . A vertex of has at most outneighbors in . It follows that has at least outneighbors in . This means that the subtournament induced by has minimum outdegree at least and then by Theorem 2.2, it contains a minimum outdegree minimal subtournament. But by maximality of , this is not possible. Consequently and therefore contains minimum outdegree minimal subtournaments. Since these tournaments are strong (Theorem 2.1), the result is proved.

Since , by Theorem 2.4, contains minimum semidegree minimal subtournaments. Let be the maximum number of vertex-disjoint minimum semidegree minimal subtournaments. Then there exist vertex-disjoint minimum semidegree minimal subtournaments, and by Theorem 2.5, these tournaments cover a set with at most vertices of . Suppose that . A vertex of has at most outneighbors in and at most in-neighbors in . It follows that has at least outneighbors in and at least in-neighbors in . This means that the subtournament induced by has minimum semidegree at least and then by Theorem 2.4, it contains a minimum semidegree minimal subtournament. But by maximality of , this is not possible. Consequently and therefore contains minimum semidegree minimal subtournaments, which proves the result.

Since the proof of Theorem 1.4 is similar to that of Theorem 1.3, we prove only Theorem 1.3.

By Theorem 2.2, contains a minimum outdegree minimal subtournament , which is strong by Theorem 2.1. By Theorem 2.3, the order of is at most . Let be the subtournament induced by . A vertex of has at least outneighbors in and at most outneighbors in . It follows that has at least outneighbors in . This means that has minimum outdegree at least , and consequently the theorem is proved.

This being established, an easy induction, gives the following.

Theorem 4.1. For an integer and for given positive integers , the vertices of any tournament with minimum outdegree at least , can be partitioned into sets so that for and is strong for .

Proof. The assertion is true for because it is Theorem 1.3. Suppose that the assertion is true up to the row , and let us study for . So, let be a tournament with minimum outdegree at least . By Theorem 1.3, can be partitioned into two sets and , so that is strong, of minimum outdegree at least and is of minimum out degree at least . By induction hypothesis can be partitioned into sets so that is of minimum outdegree at least for and is strong for . It follows, by considering also , that the assertion is true for and therefore the result is proved.

In fact, we are able to prove Conjecture 1 (and then Conjecture 2) for a larger class of oriented graphs (including tournaments). Namely, we state the following.

Theorem 4.2. For given integers , and , let be an oriented graph with minimum outdegree at least and such that for every vertex there exist at most vertices of nonadjacent with . Then the vertices of can be partitioned into two sets inducing oriented graphs and with of minimum outdegree at least , and strongly connected when , and with of minimum outdegree at least .

Proof. When , is a tournament and then the result is proved. Therefore we may suppose . By orienting every nonedge of , we obtain a tournament having as spanning subdigraph. Since has minimum out degree at least , by Theorem 1.3, can be partitioned into two sets inducing subtournaments and with of minimum outdegree at least and with of minimum outdegree at least . Since every vertex of has at most outneighbors in which are not outneighbors in , by deleting all the arcs of and which are not arcs of , we get a spanning subdigraph of , induced in by , with minimum outdegree at least and a spanning subdigraph of , induced in by and with minimum outdegree at least . Since , is a partition of , the theorem is proved.

For example an oriented graph of minimum outdegree at least 34 such that every vertex is nonadjacent with at most 4 vertices is vertexdecomposable into two oriented graphs of minimum out degree 2.

The proof of Theorem 1.4 (related to the semidegree) is similar to that of Theorem 1.3 and here also a generalization is possible.

5. An Open Problem of Thomassen

Problem 1. Let and be positive integers, does there exists a positive integer so that all but a finite number of -strong tournaments can be vertex-partitioned into an -strong and a -strong subtournament?

By minimal -strong tournament, we mean a -strong tournament such that every proper subtournament of has strongconnectivity at most . By critical -strong tournament, we mean a -strong tournament such that for every vertex , the subtournament has strong connectivity at most . We think it is as follows.

Conjecture 3. For a given integer , there exists a function such that every minimal -strong tournament is of order at most .

It is easy to prove that a positive answer to this conjecture would give a positive answer to Thomassen's open problem. It is known that the conjecture is false, when we replace minimal -strong tournament by critical -strong tournament, but in spite of that, we maintain our conjecture.